Rotational Inertia & Kinetic Energy

Linear & Angular   Linear Angular Displacement x θ Velocity v Acceleration a  Inertia m I KE ½ mv2 ½ I2 N2 F = ma  = I Momentum P = mv L = I

Rolling Motion If a round object rolls without slipping, there is a fixed relationship between the translational and rotational speeds:

Rolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion:

Rolling Motion We may also consider rolling motion at any given instant to be a pure rotation at rate w about the point of contact of the rolling object.

A Rolling Tire A car with tires of radius 32 cm drives on a highway at a speed of 55 mph. (a) What is the angular speed w of the tires? (b) What is the linear speed vtop of the top to the tires?

Rotational Kinetic Energy Consider a mass M on the end of a string being spun around in a circle with radius r and angular frequency w Mass has speed v = w r Mass has kinetic energy K = ½ M v2 K = ½ M w2 r2 Rotational Kinetic Energy is energy due to circular motion of object. M 24

Rotational Inertia I Ktran = ½ m v2 Linear Motion Tells how much “work” is required to get object spinning. Just like mass tells you how much “work” is required to get object moving. Ktran = ½ m v2 Linear Motion Krot = ½ I w2 Rotational Motion 13

Inertia Rods Two batons have equal mass and length. Which will be “easier” to spin? A) Mass on ends B) Same C) Mass in center I = S m r2 Further mass is from axis of rotation, greater moment of inertia (harder to spin)

Inertia of a Dumbbell Use the definition of moment of inertia to calculate that of a dumbbell-shaped object with two point masses m separated by a distance of 2r and rotating about a perpendicular axis through their center of symmetry. If the rod has mass then +Irod=1/12

Nose to the Grindstone A grindstone of radius r = 0.610 m is being used to sharpen an axe. If the linear speed of the stone is 1.50 m/s and the stone’s kinetic energy is 13.0 J, what is its moment of inertia I ?

Moment of Inertia of a Hoop All of the mass of a hoop is at the same distance R from the center of rotation, so its moment of inertia is the same as that of a point mass rotated at the same distance.

Moments of Inertia

I is Axis Dependent

Rolling Objects

Like a Rolling Disk A 1.20 kg disk with a radius 0f 10.0 cm rolls without slipping. The linear speed of the disk is v = 1.41 m/s. (a) Find the translational kinetic energy. (b) Find the rotational kinetic energy. (c) Find the total kinetic energy.

Preflight: Rolling Race (Hoop vs Cylinder) A hoop and a cylinder of equal mass roll down a ramp with height h. Which has greatest KE at bottom? A) Hoop B) Same C) Cylinder 20% 50% 30%

Preflight: Rolling Race (Hoop vs Cylinder) A hoop and a cylinder of equal mass roll down a ramp with height h. Which has greatest speed at the bottom of the ramp? A) Hoop B) Same C) Cylinder I = MR2 I = ½ MR2

Rolling Down an Incline

Spinning Wheel mgh=1/2mv2(1+I/mR2) A block is attached to a string around a pulley. The string is pulled rising the block at a velocity v spinning the pulley with a rotational velocity w . To what height h does the block rise? mgh=1/2mv2(1+I/mR2)

A Bowling Ball A bowling ball that has an 11 cm radius and a 5.0 kg mass is rolling without slipping at 2.0 m/s on a horizontal ball return. It continues to roll without slipping up a hill to a height h before momentarily coming to rest and then rolling back down the hill. Model the bowling ball as a uniform sphere and calculate h.

Torque and Energy Remember Torque=Fd=Iα A torque makes an acceleration So with an initial torque you can spin something up and that can roll to do work.

Example A torque of 5.0 Nm is applied to a 1.0 kg disk with a 50. cm radius for 5.0 seconds. What is the initial ω? If the coefficient of rolling friction is μ=.05 use work to find how far the wheel rolls.

Example A torque of 5.0 Nm is applied to a 1.0 kg disk with a 50. cm radius for 5.0 seconds. What is the initial ω? ω=αt τ=Iα so α= τ/I ω=t (τ/I) Idisk= 1/2mr2 ω=t (τ/(1/2mr2)) ω=5 (5/(1/2*1*.52)) = 200 rad/s2

Example A torque of 5.0 Nm is applied to a 1.0 kg disk with a 50. cm radius for 5.0 seconds. If the coefficient of rolling friction is μ=.05 use work to find how far the wheel rolls. ω=200 rad/s2 KE=1/2mv2 + 1/2Iω2 v= ωr Idisk= 1/2mr2 KE=1/2m(ωr)2 + ½(1/2mr2)ω2 KE=3/4m(ωr)2 =.75*1*(200*.5)2 = 7,500 J